Difference between revisions of "2008 USAMO Problems/Problem 2"
(I'll construct T later -_-) |
(<blank edit> Solution 1 credit to tjhance, asy by myself, Solution 2 outline credit to calc rulz, solutions 3,4 were outlined on forums but still have to write up) |
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== Solution == | == Solution == | ||
− | === Solution 1 === | + | === Solution 1 (isogonal conjugates) === |
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Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle. | Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle. | ||
− | === Solution 2 === | + | === Solution 2 (inversion) === |
+ | Invert by an arbitrary radius about A. We want to show that P', F', and N' are collinear. Notice that D', A, and P' lie on a circle with center B', and similarly for the other side. We also have that B', D', F', A form a cyclic quad and similar for the other side. We can then use some angle chasing to prove that A B' F' C' is a paralellogram, meaning that F' the midpoint of P'N'. (On the actual test, I showed that A B' F' C' was a parallelogram, and didn't realize that that implied F' was on P'N'.) {{incomplete|solution}} | ||
+ | |||
+ | === Solution 3 (trigonometric) === | ||
+ | === Solution 4 (synthetic) === | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 19:15, 1 May 2008
Problem
(Zuming Feng) Let be an acute, scalene triangle, and let , , and be the midpoints of , , and , respectively. Let the perpendicular bisectors of and intersect ray in points and respectively, and let lines and intersect in point , inside of triangle . Prove that points , , , and all lie on one circle.
Contents
[hide]Solution
Solution 1 (isogonal conjugates)
Construct on such that . Then . Then , so , or . Then , so . Then we have
and . So and are isogonally conjugate. Thus . Then
.
If is the circumcenter of then so is cyclic. Then .
Then . Then is a right triangle.
Now by the homothety centered at with ratio , is taken to and is taken to . Thus is taken to the circumcenter of and is the midpoint of , which is also the circumcenter of , so all lie on a circle.
Solution 2 (inversion)
Invert by an arbitrary radius about A. We want to show that P', F', and N' are collinear. Notice that D', A, and P' lie on a circle with center B', and similarly for the other side. We also have that B', D', F', A form a cyclic quad and similar for the other side. We can then use some angle chasing to prove that A B' F' C' is a paralellogram, meaning that F' the midpoint of P'N'. (On the actual test, I showed that A B' F' C' was a parallelogram, and didn't realize that that implied F' was on P'N'.) Template:Incomplete
Solution 3 (trigonometric)
Solution 4 (synthetic)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
2008 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>